<<

. 20
( 44 .)



>>

situation if ± and β are harmonic forms then ±§β is a harmonic form
too, but this is not true in general.

Let us introduce the following zero-order di¬erential operator
P ± = (j ’ n)±, ± ∈ Aj (M ).
¨
3.6. HODGE THEORY ON KAHLER MANIFOLDS 119

Apparently [P, L] = 2L because L increases the degree of a di¬erential
form by 2; similarly [P, Λ] = ’2Λ. Also we have
LEMMA 3.6.11
[L, Λ] = P.
Proof. Since it is a purely punctual statement, we can again use the
same arguments as in Theorem 1.14 to reduce the problem to the case
M = Cn with the standard K¨hler form. Then if we view A— (Cn ) as
a
ˆ ˆ
the completed tensor product A— (C)— · · · —A— (C), we can write

L = L1 —Id— · · · —Id + Id—L2 —Id · · · + · · ·

where Lj is the operator L on the j-th copy of C. By adjunction
we have similar decompositions for Λ and for P . Therefore we only
have to prove the statement in the case of C with the K¨hler form
a

ω = ’ ’1dz§d¯/2. Here we may check √ statement directly. On
z the

0-forms we have ΛL(1) = Λ(ω) = 2 ’1(’ ’1/2) = 1 and P (1) = ’1
as desired. On 1-forms both √ vanish identically. Finally, on 2-forms
sides
we have LΛ(dz§d¯) = L(2 ’1) = dz§d¯ and P (dz§d¯) = dz§d¯,
z z z z
thus we are done.

Now if one looks closely on the relations [L, Λ] = P , [P, L] = 2L,
and [P, Λ] = ’2Λ then one recognizes the relationship between the
operators Λ, L, and P and the algebra sl(2, R). The latter is the algebra
of traceless real 2 — 2 matrices with basis
10 01 00
H= ,X = , and Y = ,
0 ’1 00 10

which have exactly the same commutation relations: [H, X] = 2X,
[H, Y ] = ’2Y , and [X, Y ] = H. Thus the 3 dimensional Lie subalgebra
Lef spanned by Λ, L and P of the algebra Op(A— (M )) of all operators
is isomorphic to sl(2, R) via the correspondence H ’ P , X ’ L, and
Y ’ Λ. In this way the algebra A— (M ) of real-valued di¬erential forms
on M becomes a sl(2, R)-module. Because of the fact that the Laplacian
∆ commutes with L, Λ, and P one sees that the space of harmonic forms
H— (M ) ‚ A— (M ) is an sl(2, R)-submodule. Note that a vector space V
is an sl(2, R)-module if we have three operators H, X, Y on V satisfying
120 CHAPTER 3. HODGE THEORY

the above bracket relations. It follows that it is useful for us to study
¬nite-dimensional modules over sl(2, C) (also called representations of
sl(2, C)). Let V be a ¬nite-dimensional vector space acted upon by
operators X, Y and H as above. We will prove

THEOREM 3.6.12 For any integer m ≥ 0, there is exactly one (up
to isomorphism) irreducible representation Vm of sl(2, C) of dimension
m + 1.

We can describe Vm explicitly. For V0 we take just a trivial 1-dimensional
sl(2, C)-module. Let V1 = C2 with the standard action of sl(2, C), and
let Vm = S m C2 be the m-th symmetric power of V1 . The action of
sl(2, R) is extended to an action on the symmetric powers by the usual
rules for the action of a derivation on a product.
To get a di¬erent and in a way more conceptual model, let Pm
be the space of homogeneous polynomials in two variables x and y of
‚ ‚ ‚ ‚
degree m. Also, let H = x ‚x ’ y ‚y , X = x ‚y , and Y = y ‚x . It is
an easy exercise to check that we get the right commutation relations.
It is possible to ¬nd a basis in Pm consisting of eigenvectors of H,
namely, if vi = xi y m’i , then Hvi = (2i ’ m)vi for 0 ¤ i ¤ m. We have
Xvi = (m ’ i)xi+1 y m’i’1 = (m ’ i)vi+1 , and similarly Y vi = ivi’1 .
The next important result is the so-called complete reducibility of
¬nite-dimensional representations of sl(2, C).

THEOREM 3.6.13 Any ¬nite-dimensional representation V of sl(2, C)
decomposes into a direct sum

V Vm1 • · · · • Vmk

of the irreducible representations described above.

We start proving these results with a simple lemma

LEMMA 3.6.14 Let V be a ¬nite-dimensional sl(2, C)-module. If u ∈
V is such that Xu = 0, and H m u = 0 for some m ≥ 1, then Hu =
Y u = 0.
¨
3.6. HODGE THEORY ON KAHLER MANIFOLDS 121

Proof. We start with the formula

[X, Y k ] = k(H + k ’ 1)Y k’1 ,

which is easy to deduce by induction. Indeed, for k = 1 it is clear, and
if it is true for k = k0 then we have

[X, Y k0 +1 ] = [X, Y k0 ]Y + Y k0 [X, Y ] =

= k0 (H + k0 ’ 1)Y k0 + Y k0 H = [(k0 + 1)H + k0 (k0 + 1)]Y k0
as desired. Here we used that [H, Y k0 ] = ’2k0 Y k0 .
Now we can proceed with proof of the Lemma. Since H m u = 0, we
see that u ∈ V (0) . Also we notice that Y k u = 0 for some k. Assume
that k ≥ 2. The equality [X, Y k ]u = 0 and the above formula imply
that (H + k ’ 1)(Y k’1 u) = 0. It is easy to see that Y k’1 u ∈ V 2’2k ;
hoewever we also get Y k’1 u ∈ V 1’k . Because k = 1 we have Y k’1 u = 0
and we continue this process to see that Y u = 0 and therefore Hu =
[X, Y ]u = 0 as well.

Let V (») be “the generalized eigenspace” for the eigenvalue » of the
operator H. By de¬nition v ∈ V (») if and only if (H ’ »)k v = 0 for
some natural number k. A vector v ∈ V (») is said to be a vector of
generalized weight ». If furthermore we have Hv = »v we say that
v is of weight ». The space V (») could a priori be bigger than the
eigenspace V » , which is the kernel of H ’ »Id (although we will see
that in fact they are equal for ¬nite-dimensional representations). We
have the decomposition

V = •»∈C V (») .

The » such that V (») = 0 are called the weights of V . Note that X
maps V (») to V (»+2) and V » to V »+2 . Similarly Y maps V (») to V (»’2)
and V » to V »’2 . We say that a weight » is bigger than a weight µ if
» ’ µ ∈ {1, 2, 3 · · ·}. We will soon prove that any weight is always an
integer. If V is a ¬nite-dimensional representation it contains a vector
v of maximal weight ». Then we must have X · v = 0. Let v be a vector
of maximal weight » in V . We claim that » must be an integer. Indeed
let k be the smallest integer such that Y k · v = 0. Then v is killed by
122 CHAPTER 3. HODGE THEORY

[X, Y k ] = k(H + k ’ 1)Y k’1 . Since Y k’1 v is of weight » ’ 2k + 2, it
follows that (H + k ’ 1)Y k’1 · v = (» ’ k + 1)Y k’1 v. Therefore we have
» = k ’ 1. We claim that if V is irreducible then H is diagonalizable,
i.e., V (») = V » for all ». Indeed the eigenvectors of H span an sl(2, C)-
submodule which is non-trivial, hence must be equal to V . It is then
easy to construct a linear map φ : Vk’1 ’ V which sends v0 to v, and
which commutes with the operators H and Y . We then observe

LEMMA 3.6.15 φ is an sl(2, C)-linear map.

Proof. This follows immediately from Lemma 3.6.14.
This proves Theorem 3.6.12.
Now given any ¬nite-dimensional representation V we can ¬nd a
Jordan-H¨lder series
o

0 ‚ E1 ‚ · · · ‚ Ek ‚ V,

where E1 , ..., Ek are sl(2, C)-submodules, such that this ¬ltration is pre-
served by H, X, and Y ; and E1 , E2 /E1 , ..., V /Ek all are irreducible
representations. The di¬culty is to show that E2 ’E1 •(E2 /E1 ), etc...
˜

LEMMA 3.6.16 Let

0’V ’W ’C’0

be an extension of sl(2, C) ¬nite-dimensional representation V . Then
W = V • C as an sl(2, C)-representation.

Proof. We observe that X : V (0) ’ V (2) is surjective, for irreducible
representations. It follows that the same is true for any ¬nite-dimensional
representation, because given an exact sequence 0 ’ E ’ F ’ G ’ 0
of sl(2, C)-representations, we have a diagram with exact rows:

0 ’ E (0) ’ F (0) ’ G(0) ’ 0
“X “X “X .
0 ’ E (2) ’ F (2) ’ G(2) ’ 0
¨
3.6. HODGE THEORY ON KAHLER MANIFOLDS 123

We have
onto
Ker(X on W ) ’ C
“ “
0 ’ V (0) W (0)
’ ’ ’0
C
“X “X “X
0 ’ V (2) W (2)
’ ’ 0 ’0
“ “
0 0

The fact that the kernel of the mapping X : W (0) ’ W (2) maps onto C
means that there exists u ∈ W such that Xu = 0, u ∈ W (0) , and u maps
to 1 ∈ C. By the previous Lemma we have that u generates a trivial
1-dimensional sl(2, C) representation C · u, and therefore W = V • C · u.

We are now ready to prove
PROPOSITION 3.6.17 Let 0 ’ V1 ’ V2 ’ V3 ’ 0 be an ex-
act sequence of ¬nite-dimensional representations of sl(2, C). Then the
extension is split, i.e., V3 ’V1 • V2 as sl(2, C)-representations.
˜
Proof. Given representations A and B of sl(2, C), the space Hom(A, B)
becomes a representation by the rule (Xφ)(a) = X · φ(a) ’ φ(X · a) for
φ ∈ Hom(A, B), a ∈ A. Then the space E of linear maps f : V2 ’ V1
such that f (v) = »v for all v ∈ V1 is a subrepresentation of Hom(V2 , V1 )
and we have an exact sequence
»
0 ’ Hom(V3 , V1 ) ’ E ’ C ’ 0.

The previous lemma implies that the extension splits. This means that
there exists an element f of E which is killed by sl(2, C) and induces
the identity on V1 ‚ V2 . The fact that u is killed by sl(2, C) precisely
means that u is a homomorphism of sl(2, C)-representations. Then u is
onto, and if K is its kernel we have a decomposition V2 = V1 • K of
sl(2, C)-representations, where K ’V3 . This proves the splitting.
˜
Now starting with a Jordan-H¨lder series 0 ‚ E1 ‚ · · · ‚ Ek ‚ V
o
for a representation V we see inductively that Ej is a direct sum of
irreducible representations. This proves Theorem 3.6.13. As another
consequence one obtains
124 CHAPTER 3. HODGE THEORY

COROLLARY 3.6.18 For any ¬nite-dimensional representation V
of
sl(2, C) the operator H is diagonalizable and the map

X j : V ’j ’ V j , j > 0

is an isomorphism.

Proof. We can assume that V is an irreducible representation of sl(2, C):
V Vm . Then using the explicit structure of Vm = Pm we see that
when j > m then V ’j = V j = 0 and when j ¤ m the action of
m’j m+j

X j = (x ‚y )j is non zero on the monomial x 2 y 2 in Pm .

Since H is diagonalizable, we also point out that V (») = V » .



THEOREM 3.6.19 (Hard Lefschetz theorem.) Let M be a com-
pact K¨hler manifold of dimension n. Then the map
a

Lk : H n’k (M, C) ’ H n+k (M, C), k ≥ 0

is an isomorphism.

Proof. We use the isomorphism H — (M, C) H— (M ) together with the
correspondence X ” L, Y ” Λ, and H ” P to view H— (M ) as a
representation space of sl(2, C). On (n ’ k)-forms P acts as ’k and
thus the result follows from the following commutative diagram.

Lk
n’k
(M ) ’’ Hn+k (M )
H
|| ||
Xk
V ’k Vk
’’



COROLLARY 3.6.20 If p is odd then dim H p (M, R) is an even num-
ber.
¨
3.6. HODGE THEORY ON KAHLER MANIFOLDS 125

Proof. We will give two proofs. The ¬rst is simple and the second is

<<

. 20
( 44 .)



>>